$ D = \left[\begin{array}{rrr}-1 & -1 & 4 \\ 0 & 5 & 0\end{array}\right]$ $ F = \left[\begin{array}{rr}1 & 4 \\ 2 & 0 \\ 1 & 5\end{array}\right]$ What is $ D F$ ?
Solution: Because $ D$ has dimensions $(2\times3)$ and $ F$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D F = \left[\begin{array}{rrr}{-1} & {-1} & {4} \\ {0} & {5} & {0}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{4} \\ {2} & \color{#DF0030}{0} \\ {1} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{1}+{-1}\cdot{2}+{4}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{1}+{-1}\cdot{2}+{4}\cdot{1} & ? \\ {0}\cdot{1}+{5}\cdot{2}+{0}\cdot{1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{1}+{-1}\cdot{2}+{4}\cdot{1} & {-1}\cdot\color{#DF0030}{4}+{-1}\cdot\color{#DF0030}{0}+{4}\cdot\color{#DF0030}{5} \\ {0}\cdot{1}+{5}\cdot{2}+{0}\cdot{1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{1}+{-1}\cdot{2}+{4}\cdot{1} & {-1}\cdot\color{#DF0030}{4}+{-1}\cdot\color{#DF0030}{0}+{4}\cdot\color{#DF0030}{5} \\ {0}\cdot{1}+{5}\cdot{2}+{0}\cdot{1} & {0}\cdot\color{#DF0030}{4}+{5}\cdot\color{#DF0030}{0}+{0}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}1 & 16 \\ 10 & 0\end{array}\right] $